Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		double a[]=new double[n];
		for(int i=0;i<n;i++){
			a[i]=sc.nextDouble();
		}
		double dp[][]=new double[n+1][n+1];
		dp[0][0]=1;
		double count=0;
		for(int h=0;h<=n;h++)
		for(int t=0;t<=n;t++){
			if(h+t>n)
			continue;
			if(h>0){
				dp[h][t]+=dp[h-1][t]*a[h+t-1];
			}
			if(t>0){
				dp[h][t]+=dp[h][t-1]*(1-a[h+t-1]);
			}
			if(h+t==n&&h>t){
				//pw.println(h+" "+t+" "+dp[h][t]);
				count+=dp[h][t];
			}
		}
		System.out.println(count);

	}
}